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The sum of the co-efficients of all the even powers of x in the expansion of (2x2 – 3x + 1)11 is
- a)2.610
- b)3.610
- c)611
- d)None
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
The sum of the co-efficients of all the even powers of x in the expans...
The sum of the coefficients of all the even powers of x in the expansion of (2x^2 - 3x + 1)^11 can be found by multiplying the expansion by x^2 and collecting the terms that contain even powers of x.
The expansion of (2x^2 - 3x + 1)^11 is given by:
(2x^2 - 3x + 1)^11 = 2^11 * x^22 + 11 * 2^10 * x^4 - 55 * 2^9 * x^6 + 220 * 2^8 * x^8 + ...
Multiplying this expansion by x^2 gives:
x^2 * (2x^2 - 3x + 1)^11 = 2^11 * x^4 + 11 * 2^10 * x^6 - 55 * 2^9 * x^8 + 220 * 2^8 * x^10 + ...
The sum of the coefficients of all the even powers of x in this expansion is given by:
2^11 + 11 * 2^10 - 55 * 2^9 + 220 * 2^8 = 2048 + 121,344 - 117,760 + 8,704,000 = 811,832
Therefore, the sum of the coefficients of all the even powers of x in the expansion of (2x^2 - 3x + 1)^11 is 811,832.
The correct answer is therefore (a) 2.610.
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Community Answer
The sum of the co-efficients of all the even powers of x in the expans...
To find the sum of the coefficients of all the even powers of x in the expansion of (2x^2)^n, we can use the Binomial Theorem.
The Binomial Theorem states that for any positive integer n and any real numbers a and b:
(a + b)^n = C(n,0) * a^n * b^0 + C(n,1) * a^(n-1) * b^1 + C(n,2) * a^(n-2) * b^2 + ... + C(n,n-1) * a^1 * b^(n-1) + C(n,n) * a^0 * b^n
where C(n, k) = n! / (k! * (n-k)!) is the binomial coefficient.
In this case, a = 2x^2 and b = 0, so the expansion becomes:
(2x^2)^n = C(n,0) * (2x^2)^n * 0^0 + C(n,1) * (2x^2)^(n-1) * 0^1 + C(n,2) * (2x^2)^(n-2) * 0^2 + ... + C(n,n-1) * (2x^2)^1 * 0^(n-1) + C(n,n) * (2x^2)^0 * 0^n
Since any term with a power of 0 will be 0, we can ignore those terms. Therefore, the sum of the coefficients of all the even powers of x is equal to the sum of the coefficients of the terms with powers of x^2, x^4, x^6, and so on.
To find the sum of the coefficients of these terms, we need to substitute x^2 = 1 into the expansion. This means that in each term, we can replace x^2 with 1, and simplify the expression.
(2x^2)^n = C(n,0) * (2x^2)^n * 0^0 + C(n,1) * (2x^2)^(n-1) * 0^1 + C(n,2) * (2x^2)^(n-2) * 0^2 + ... + C(n,n-1) * (2x^2)^1 * 0^(n-1) + C(n,n) * (2x^2)^0 * 0^n
= C(n,0) * (2 * 1)^n + C(n,1) * (2 * 1)^(n-1) + C(n,2) * (2 * 1)^(n-2) + ... + C(n,n-1) * (2 * 1)^1 + C(n,n) * (2 * 1)^0
= C(n,0) * 2^n + C(n,1) * 2^(n-1) + C(n,2) * 2^(n-2) + ... + C(n,n-1) * 2^1 + C(n,n) * 2^0
Now we can use the binomial coefficient formula to calculate each term. The binomial coefficient C(n, k) is equal to n!
The Binomial Theorem states that for any positive integer n and any real numbers a and b:
(a + b)^n = C(n,0) * a^n * b^0 + C(n,1) * a^(n-1) * b^1 + C(n,2) * a^(n-2) * b^2 + ... + C(n,n-1) * a^1 * b^(n-1) + C(n,n) * a^0 * b^n
where C(n, k) = n! / (k! * (n-k)!) is the binomial coefficient.
In this case, a = 2x^2 and b = 0, so the expansion becomes:
(2x^2)^n = C(n,0) * (2x^2)^n * 0^0 + C(n,1) * (2x^2)^(n-1) * 0^1 + C(n,2) * (2x^2)^(n-2) * 0^2 + ... + C(n,n-1) * (2x^2)^1 * 0^(n-1) + C(n,n) * (2x^2)^0 * 0^n
Since any term with a power of 0 will be 0, we can ignore those terms. Therefore, the sum of the coefficients of all the even powers of x is equal to the sum of the coefficients of the terms with powers of x^2, x^4, x^6, and so on.
To find the sum of the coefficients of these terms, we need to substitute x^2 = 1 into the expansion. This means that in each term, we can replace x^2 with 1, and simplify the expression.
(2x^2)^n = C(n,0) * (2x^2)^n * 0^0 + C(n,1) * (2x^2)^(n-1) * 0^1 + C(n,2) * (2x^2)^(n-2) * 0^2 + ... + C(n,n-1) * (2x^2)^1 * 0^(n-1) + C(n,n) * (2x^2)^0 * 0^n
= C(n,0) * (2 * 1)^n + C(n,1) * (2 * 1)^(n-1) + C(n,2) * (2 * 1)^(n-2) + ... + C(n,n-1) * (2 * 1)^1 + C(n,n) * (2 * 1)^0
= C(n,0) * 2^n + C(n,1) * 2^(n-1) + C(n,2) * 2^(n-2) + ... + C(n,n-1) * 2^1 + C(n,n) * 2^0
Now we can use the binomial coefficient formula to calculate each term. The binomial coefficient C(n, k) is equal to n!
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The sum of the co-efficients of all the even powers of x in the expansion of (2x2–3x + 1)11isa)2.610b)3.610c)611d)NoneCorrect answer is option 'B'. Can you explain this answer?
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